Volume of a Restricted Region in n-Dimensional Space

In summary, the volume of {x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ n} is n^n/(n!)^2.
  • #1
Frillth
80
0

Homework Statement



Let S = {x∈R^n : x_i ≥ 0 for all i, x_1 + 2x_2 + 3x_3 + ... + nx_n ≤ n}. Find the n-dimensional volume of S.

Homework Equations



I'm 95% sure that I'm supposed to use the change of variables theorem here.

The Attempt at a Solution



So far, I have calculated the values for n from 1 to 6 using iterated integrals (calculated with mathematica). They are 1, 1, 3/4, 4/9, 125/576, and 9/100, respectively. I was hoping to get an easy pattern that I could use to lead me toward the general solution, but it looks like that won't be the case.

This is only the second COVT problem that we've been assigned, so I'm not quite comfortable with the theorem yet. I don't really have any idea what sort of change of variables would make this easy to deal with. Does anyone have any idea?

Edit: I just figured out the pattern based on some intuition and a lucky guess. The volume of S is (or seems to be, at least) n^n/(n!)^2. I have no idea how I could possibly get this by integration, however.
 
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  • #2
Frillth said:
Let S = {x∈R^n : x_i ≥ 0 for all i, x_1 + 2x_2 + 3x_3 + ... + nx_n ≤ n}. Find the n-dimensional volume of S.

Edit: I just figured out the pattern based on some intuition and a lucky guess. The volume of S is (or seems to be, at least) n^n/(n!)^2. I have no idea how I could possibly get this by integration, however.

Hi Frillth! :smile:

Hint: what is the volume of {x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ n}? :wink:
 
  • #3
The volume of that should be (I think):
sw7rc7.png

Is there a simple way to evaluate this, or am I doing this completely wrong?

By the way, sorry if that's hard to read. I don't know latex, so I just used Mathematica.

Edit: Ah, I just realized how to compute that integral. It comes out to n^n/n!, right?
Edit 2: So for my change of variables to, say, u, I want u_1 = x_1, u_2 = 2x_2, u_3 = 3x_3 ... u_n = nx_n. Then the determinant of the derivative matrix for this function would be 1/n!. So I take the integral of the surface described by u_1 + u_2 + ... u_n = n, which I already showed was n^n/n!, and multiply by 1/n! from the determinant to get volume = n^n/(n!)^2. Is that correct?
Edit 3: All right, I believe I have a rigorous proof now. Thank you very much for your hint, tiny-tim.
 
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  • #4
Frillth said:
Ah, I just realized how to compute that integral. It comes out to n^n/n!, right?

Hi Frillth! :smile:

(for LaTeX integrals, type \int_{a}^{b} and for three dots type \cdots :wink:)

That was fun! :smile:

Which way did you get the n^n/n!?

I found two ways (I did it for a "pyramid" of side 1, so I just got 1/n!) …

one was ordinary integrating, by induction on n,

and the other was by comparing

{x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ 1}

with

{x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ 1

and x_1 ≤ x_2 ≤ x_3 ≤ ... ≤ x_n} :wink:

As a result of more geometric juggling, I also managed to prove:

[tex]\sum_{0\,\leq\,m\,< n/2}\,(-1)^m(n - 2m)^n\,^nC_m\ =\ 2^{n-1}\,n![/tex]

… see https://www.physicsforums.com/showthread.php?p=2083077". :smile:
 
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  • #5
I did it by induction on n. How would you approach it with your second method?
 
  • #6
Hi Frillth! :smile:
Frillth said:
I did it by induction on n. How would you approach it with your second method?

waaah! :cry: I've just spent ages trying to re-create it because i couldn't read my own handwriting! :redface:

that second set should have been …

{y∈R^n : 0≤ y_1 ≤ y_2 ≤ y_3 ≤ ... ≤ y_n ≤ 1}

with y_m = x_i + … + x_m :wink:

then the Jacobian is 1, and the original integral equals

[tex]\int_0^1\int_0^{y_n}\cdots\int_0^{y_2}dy_n\cdots dy_1[/tex]

= 1/n! :rolleyes:

neat, huh? :wink:

btw, this technique, of ordering the variables, is used a lot in quantum field theory, where each y is a time, and the integral is called a time-ordered integral

now it's your turn …

i've only just noticed that there's a direct proof which is (obviously :rolleyes:) what they intended you to use …

can you see a similar method, using a change of variables with a Jacobian of n!, which directly solves for the original {x∈R^n : x_i ≥ 0 for all i, x_1 + 2x_2 + 3x_3 + ... + nx_n ≤ n}? :smile:
 

Related to Volume of a Restricted Region in n-Dimensional Space

1. What is the Change of Variables Theorem?

The Change of Variables Theorem is a mathematical theorem that describes how to change the variables in a double or triple integral. It states that if a function can be expressed in terms of a different set of variables, then the integral of that function over a given region can also be expressed in terms of the new variables.

2. What is the purpose of the Change of Variables Theorem?

The purpose of the Change of Variables Theorem is to simplify the evaluation of integrals by allowing us to change the variables to ones that are easier to integrate. This can make complicated integrals more manageable and can also help us to solve problems that would otherwise be impossible to solve using traditional methods.

3. How is the Change of Variables Theorem used in real-life applications?

The Change of Variables Theorem has many real-life applications, especially in physics and engineering. It is used to solve problems involving multiple variables, such as calculating the center of mass of an object or determining the flow of fluid through a complex system. It is also used in statistical analysis and in solving differential equations.

4. What are the limitations of the Change of Variables Theorem?

The Change of Variables Theorem can only be applied to functions that are continuous and have continuous derivatives. It also requires that the change of variables is a one-to-one transformation. Additionally, the region of integration must be transformed along with the variables, which can be challenging in some cases.

5. How can I remember the formula for the Change of Variables Theorem?

The formula for the Change of Variables Theorem can be remembered using the mnemonic "U-substitution." This stands for "u equals some function," which represents the new variables in the integral. Additionally, practicing with different examples and understanding the concept behind the theorem can help with remembering the formula.

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